Continuous functions in the indiscrete topology? Slight curiosity. I've learned not to question too much in topology and basically, acquiesce. In a sense, thinking hard or trying to be smart in this area of study is a suicide mission for newbies. Unless for uber futuristic geniuses of course.
Anyway, so a continuous function $f:X \to Y$, in topology is defined as

$f^{-1}:Y \to X$ maps open sets to open sets.

Sure. I don't see how this relates to previously learned continuity of "the graph can be drawn without lifting the pen" or the more rigorous definition using $\lim f(x)$. So this isn't my question.
My question is, well, if we are equipping $X$ with the indescrete topology $\tau=\{X,\phi\}$ then... does any $f:X \to Y$ become continuous?
Well, in this topology, the only open sets are the entire set and the null set. In other words, I take any element $x \in X$ and that element is... wait, as I am writing this, I have another question; is this $x$ open in $X$? My confusion is, when we say open in $X$ we often talk about "sets" or "subsets." Here, Iv've hand picked an element and asked if it is open or not. $X$ itself is open, (as a set) but then, are each elements of it also open too? Or do I have to take all $x$'s in $X$ collectively?
Okay, that's one question, and so back to my original query, say if $x$ is open in $X$, then,  whatever I take from $Y$, say $y \in Y$, the inverse $f^{-1}$ will map to some element in $X$...which is an element of an open set $X$. Whatever way I define $f$, as long as there is an inverse, in this case, is $f$ is continuous regardless?
My actual question is dependent on the answer to my spontaneous question but can someone clear these up for me?
 A: First, usually there is no map $f^{-1}:Y\to X$, what we have is a map $f^{-1} : P(Y) \to P(X)$ defined by $f^{-1}(E) = \{ x\in X | f(x) \in E \}$. And this is always well defined.
About your first question :
If you are equiping $X$ with the indiscrete topology, then you don't have continuity in most cases : if your topology on $Y$ is separated, the only continuous functions are constant. Indeed, if $f$ take two values $a$ and $b$, as $Y$ is separated, there is an open set $U$ that contain $a$ but not $b$, that means that $f^{-1}(U) \neq \emptyset$ (because there is an x that verify $f(x) = a$) and $f^{-1}(U) \neq X$ (because there an x that verify $f(x) = b \not\in U$) Hence $f^{-1}(U)$ is not open
A: First, the open sets in $X$ are by definition subsets of $X$; if $x\in X$, it makes no sense to ask whether $x$ is open. It is meaningful to ask whether the set $\{x\}$ is open, however.
Now suppose that $X$ is given the indiscrete topology, so that the only open subsets of $X$ are $\varnothing$ and $X$ itself, and let $f:X\to Y$. Then $f$ is continuous if and only if the following statement is true:

whenever $U$ is an open subset of $Y$, $f^{-1}[U]$ is an open subset of $X$.

Since the only open subsets of $X$ are $\varnothing$ and $X$, in this special case we can rewrite the condition:

whenever $U$ is an open subset of $Y$, $f^{-1}[U]$ is either empty or $X$.

Suppose that $U\subseteq Y$, and $f^{-1}[U]=\varnothing$; then $U\cap f[X]=\varnothing$. In other words, $U$ is disjoint from the range of $f$. Now suppose that $f^{-1}[U]=X$; then $U\supseteq f[X]$. In other words, $f$ is continuous if every open subset of $Y$ either contains or is disjoint from $f[X]$.
In general this will not be the case. Suppose, for example, that $Y$ is $T_0$, and $f[X]$ contains at least two points. Let $y_0$ and $y_1$ be distinct points of $f[X]$. Since $Y$ is $T_0$, either there is an open set $U$ in $Y$ such that $y_0\in U$ and $y_1\notin U$, or there is an open set $U$ in $Y$ such that $y_1\in U$ and $y_0\notin U$. In either case $U$ is not disjoint from $f[X]$, since it contains one of the points $y_0$ and $y_1$, but it also does not contain $f[X]$, since it fails to contain one of the points $y_0$ and $y_1$. Thus, $f^{-1}[U]$ is neither $\varnothing$ nor $X$, and $f$ is not continuous.

Matters are very different if $X$ has the discrete topology. Then every subset of $X$ is open, so $f^{-1}[U]$ is open in $X$ for every subset $U$ of $Y$, open or not, and $f$ is therefore automatically continuous.
A: $f^{-1}$ is not necessarily a function so $f^{-1}:Y \rightarrow X$ doesn't always make sense, but you can talk about pre-images of open sets under $f$.  Also, points of $X$ are not technically subsets of $X$, so you shouldn't talk about $x \in X$ being open.  You can ask if $\{x\} \subset X$ is open, which it isn't in the indiscrete topology unless $\{x\} = X$; as you stated the only opens are $X$ and $\varnothing$.
It's not true that all $f:X \rightarrow Y$ are continuous when X has the indiscrete topology. Take $X=Y=\{1,2,3\}$ with the indiscrete topology on X, and the discrete topology on Y (all subsets are open).  Then the identity function $f(x) = x$ is not continuous since $f^{-1}(\{2\}) = \{2\}$ which is not open in $X$.
